Problem Sets

These problems combine results that are often learned separately. Each solution identifies which theorem is doing the work, then carries out a calculation or checks a limiting claim. They supplement the exercises in the linked articles; they do not replace the hypotheses and proofs given there.

Helpful background. Use Born probabilities from spectral measures, trace distance, and complete positivity of dynamics.

Spectral distributions of an additive observable

On $\mathcal H=L^2([0,1]^2,dx\,dy)$, let $H$ multiply by $x+y$ and let $\psi(x,y)=1$. All quantities in this problem are dimensionless.

1. Find the spectrum and determine whether $H$ has eigenvectors.
2. Compute the probability density of an energy measurement in $\psi$, its mean, and its variance.
3. Find $\Pr(H\leq1)$ and explain why its nonzero value does not imply that $1$ is an eigenvalue.

Solution

The operator is bounded and self-adjoint on all of $\mathcal H$. Its spectrum is the essential range $[0,2]$. An eigenvector at $\lambda$ would be supported on the line $x+y=\lambda$, a set of two-dimensional Lebesgue measure zero. Hence every eigenspace is zero.

The state is normalized because the unit square has area one. The spectral law is the pushforward of uniform area under $(x,y)\mapsto x+y$. Integrating along the allowed interval of $x$ gives

$$f(E)= \begin{cases} E,&0<E<1,\\ 2-E,&1<E<2,\\ 0,&\text{otherwise}. \end{cases}$$

One may check normalization by adding the areas of the two triangles. Equivalently, the two commuting coordinate observables have independent uniform laws in this product state. Therefore

$$\langle H\rangle=\frac12+\frac12=1,\qquad \operatorname{Var}(H)=\frac1{12}+\frac1{12}=\frac16.$$

The lower half of the square has area $1/2$, so $\Pr(H\leq1)=1/2$, whereas $\Pr(H=1)=0$. A spectral projection for an interval can be nonzero even when every singleton projection vanishes. The appropriate theorem is the joint spectral measure, not a sum over a discrete energy basis.

A quantitative thermal cutoff

Let $H|n\rangle=n^2|n\rangle$ on $\ell^2(\mathbb N_0)$, with its spectral domain. Fix $\beta>0$ and define

$$Z=\sum_{n=0}^\infty e^{-\beta n^2},\qquad \rho_\beta=Z^{-1}\sum_{n=0}^\infty e^{-\beta n^2}|n\rangle\langle n|.$$

Let $\rho_{\beta,N}$ be the normalized restriction to $0\leq n\leq N$.

1. Prove that $Z$ is finite and bound the omitted partition sum.
2. Compute the trace distance $D(\rho_\beta,\rho_{\beta,N})$ exactly in terms of the omitted Gibbs probability.
3. Obtain an explicit sufficient condition for every effect probability to differ by at most $\varepsilon$.

Solution

For $n\geq1$, $n^2\geq n$, so a geometric series proves convergence. Put

$$R_N=\sum_{n=N+1}^\infty e^{-\beta n^2},\qquad \delta_N=\frac{R_N}{Z}.$$

Successive terms in this tail have ratio $e^{-\beta(2n+1)}\leq e^{-\beta(2N+3)}$. Consequently

$$R_N\leq \frac{e^{-\beta(N+1)^2}}{1-e^{-\beta(2N+3)}}.$$

On the retained subspace the normalized cutoff increases every probability by the common factor $(1-\delta_N)^{-1}$. The total increase is $\delta_N$; the discarded tail also has total mass $\delta_N$. The two states commute, so their trace-norm difference is the sum of absolute diagonal differences:

$$\|\rho_\beta-\rho_{\beta,N}\|_1=2\delta_N,\qquad D(\rho_\beta,\rho_{\beta,N})=\delta_N.$$

Since $Z\geq1$, requiring the displayed upper bound on $R_N$ to be at most $\varepsilon$ is sufficient. The operational characterization of trace distance then bounds the error for every effect $0\leq E\leq I$ by $\varepsilon$. This is a uniform probability statement. It is not an automatic error estimate for the unbounded energy observable; that requires controlling a tail with an additional factor $n^2$.

Small state error with large energy error

Let $N|n\rangle=n|n\rangle$ on $\ell^2(\mathbb N_0)$ and, for integers $k\geq2$, set

$$\sigma_k= \left(1-\frac1k\right)|0\rangle\langle0| +\frac1k|k^2\rangle\langle k^2|.$$

Find the trace distance to the vacuum, the first and second number moments, and the difference in expectations of the bounded resolvent $(N+I)^{-1}$. Decide whether trace-norm convergence implies convergence of the mean energy.

Solution

The diagonal difference has two nonzero entries, $-1/k$ and $1/k$. Thus

$$D(\sigma_k,|0\rangle\langle0|)=\frac1k,\qquad \operatorname{Tr}(\sigma_kN)=k,\qquad \operatorname{Tr}(\sigma_kN^2)=k^3.$$

Each state has finite moments, but the first moments are not uniformly bounded. The resolvent expectation is

$$\operatorname{Tr}\bigl(\sigma_k(N+I)^{-1}\bigr) =1-\frac1k+\frac1{k(k^2+1)}.$$

It tends to the vacuum value $1$, with absolute difference $k/(k^2+1)$. There is no contradiction: trace-norm convergence controls every fixed bounded observable, while $N$ is unbounded. The relevant missing property for energy convergence is a suitable uniform integrability condition on the energy tails. Expectation Functionals distinguishes a finite expectation from a uniform estimate along a family.

In particular, replacing an unbounded observable by a bounded function of it changes which convergence claim is being tested. A numerically stable resolvent expectation does not certify convergence of mean energy.

Bounded jump dynamics without an equilibrium state

Let $S|n\rangle=|n+1\rangle$ be the unilateral shift and $\gamma>0$. The Poissonized channel construction gives

$$\Phi_t(X)=e^{-\gamma t} \sum_{j=0}^\infty\frac{(\gamma t)^j}{j!} S^jX(S^\dagger)^j.$$

The series converges in trace norm. Use this channel semigroup to answer:

1. What is the generator and why is it bounded on trace class?
2. Starting in the vacuum, what are the number distribution, mean, and variance?
3. Can any normal state be stationary?

Solution

The map $C(X)=SXS^\dagger$ is trace preserving because $S^\dagger S=I$. The canonical Poissonized-channel argument applies and gives

$$\mathcal L=\gamma(C-\mathrm{id}),\qquad \|\mathcal L\|_{1\to1}\leq2\gamma.$$

For vacuum input, the displayed expansion is diagonal with probabilities $p_n(t)=e^{-\gamma t}(\gamma t)^n/n!$. Hence both the mean and variance of $N$ equal $\gamma t$. This moment calculation uses the convergent Poisson sums, not boundedness of $N$.

If a normal stationary state $\rho$ existed, write $p_n=\langle n|\rho|n\rangle$. Stationarity implies $\mathcal L(\rho)=0$, whose diagonal equations are

$$0=-\gamma p_0,\qquad 0=\gamma(p_{n-1}-p_n)\quad(n\geq1).$$

Induction gives $p_n=0$ for every $n$, contradicting $\sum_np_n=1$. Off-diagonal matrix elements cannot repair this normalization failure. Complete positivity, conservativity, and a bounded generator do not imply a normal equilibrium state. Probability moves to arbitrarily high levels while total probability remains one.

References

• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Spectral calculus and tensor observables.
• Holevo, Alexander S. Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd ed. De Gruyter, 2019. doi:10.1515/9783110642490. Normal states, trace distance, and completely positive dynamics.